Probability Question~?
2017-05-10 1:21 am
I need help. A list of numbers contains n odd numbers and m even numbers. If a number is picked at random from the list, the probability that an odd number is selected is 2/7. What is the value of n in terms of m?
回答 (3)
2017-05-10 1:25 am
In (m + n) numbers, there are n odd numbers.
P(an odd number) :
n/(m + n) = 2/7
7n = 2(m + n)
7n = 2m + 2n
5n = 2m
n = 2m/5
P(an odd number) :
n/(m + n) = 2/7
7n = 2(m + n)
7n = 2m + 2n
5n = 2m
n = 2m/5
2017-05-10 1:57 am
Totally, there are m+n numbers.
P( odd number) = n/(n+1)
n/(m+n) = 2/7
cross multiply
7n = (m+n)(2)
7n = 2m+2n
5n = 2m
n = (2/5) m -----> n in terms of m
P( odd number) = n/(n+1)
n/(m+n) = 2/7
cross multiply
7n = (m+n)(2)
7n = 2m+2n
5n = 2m
n = (2/5) m -----> n in terms of m
2017-05-10 1:35 am
2/7 = n/(n + m)
7n = 2n + 2m
5n = 2m
n = 2m/5
7n = 2n + 2m
5n = 2m
n = 2m/5
收錄日期: 2021-04-18 16:10:48
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